Question

Find the angle between the curves

x + y + 2 = 0; x^{2} + y^{2} - 10 y = 0

Accepted Answer

Given equation of line is $x + y + 2 = 0$

$\Rightarrow x = - (y + 2)$ ......(i)

Equation of the curve $x^{2} + y^{2} - 10 y = 0$ ......(ii)

Solving equation (i) and (ii), we get

$(y + 2)^{2} + y^{2} - 10 y = 0$

$\Rightarrow y^{2} + 4 y + 4 + y^{2} - 10 y = 0$

$\Rightarrow 2 y^{2} - 6 y + 4 = 0$

$\Rightarrow y^{2} - 3 y + 2 = 0$

$\Rightarrow (y + 1) (y - 2) = 0$

$\Rightarrow y = 1 or y = 2$

When $y = 1$ ,

$\Rightarrow x = - (1 + 2) = - 3$ (from equation(i))

When $y = 2$ ,

$\Rightarrow x = - (2 + 2) = - 4$ (from equation(i))

The points of intersection are $P (- 3, 1)$ and $Q (- 4, 2)$

Now, differentiate equation(ii) w.r.to $x$ .

$\Rightarrow 2 x + 2 y \frac{d y}{d x} - 10 \frac{d y}{d x} = 0$

$\Rightarrow 2 \frac{dy}{dx} (y - 5) = - 2 x$

$\Rightarrow \frac{dy}{dx} = - \frac{x}{y - 5}$

From equation (i), compare with the $y = m x + c$

Slope is $m_{2} = - 1$

Case(i):

Slope $m_{1}$ $= \frac{d y}{d x}$ at $P$ $= - \frac{- 3}{1 - 5} = - \frac{3}{4}$ and slope $m_{2} = - 1$ .

Let $θ$ be the angle between the curves, then $tanθ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}|$

$= |\frac{- \frac{3}{4} + 1}{1 + \frac{3}{4}}| = \frac{1}{7}$

Case(ii):

Slope $m_{1}$ $= \frac{d y}{d x}$ at $Q$ $- \frac{4}{2 - 5} = - \frac{4}{3}$ and slope $m_{2} = - 1$ .

Let $θ$ be the angle between the curves, then $tanθ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}|$

$= |\frac{- \frac{4}{3} + 1}{1 + \frac{4}{3}}| = \frac{1}{7}$

$∴ θ = \tan^{- 1} (\frac{1}{7})$

Andhra Pradesh Board Solutions for Chapter: Applications of Derivatives, Exercise 4: Exercise 10(d)

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Chapter: Applications of Derivatives, Exercise 4: Exercise 10(d)

Questions from Andhra Pradesh Board Solutions for Chapter: Applications of Derivatives, Exercise 4: Exercise 10(d) with Hints & Solutions

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